Menu

Edexcel GCSE Computer Science (1CP2) Exam Questions

Build a practice paper from 693 exam-style questions, each with its mark scheme and example answers. Print it, or answer it on paper and then open the mark schemes. Or browse the questions by topic.

Build a paper

Topics (none ticked: all of them)
Question types (none ticked: all of them)

Practice paper

2 questions · 2 marks

Pick again
Print
1Multiple choice1 markSound Swap Remove

A sound file is created by sampling an analogue sound wave.

Identify the term for the number of samples taken each second.

[1 mark]
  • ABit depth
  • BSample interval
  • CSample rate
  • DSample resolution
Mark scheme
  • C Sample rate (1)
  • The only correct answer is C. The correct answer is sample rate.
  • A is not correct because bit depth is the number of bits used to record each sample
  • B is not correct because the sample interval is the time between each sample
  • D is not correct because sample resolution is another name for the bit depth
Example answers and how they were marked
Full marks1/1

C Sample rate

Why: Selects the only correct option (1).

No marks0/1

B Sample interval

Why: Near miss: the sample interval is the time between samples, not the number of samples per second (0).

No marks0/1

D Sample resolution

Why: Misconception reported in examiners' reports: sample resolution means the bit depth, not the number of samples per second (0).

2Short recall1 markCharacter Sets Swap Remove

A computer can store a maximum of 80 bits for a text message. Each character is stored using 7-bit ASCII.

State the maximum number of whole characters that can be stored in the message.

[1 mark]
Mark scheme

Award one mark for:

  • 11 (characters) (1)

Allow 11 remainder 3.

Allow 80 ÷ 7 = 11.428... rounded down to 11.

Do not accept 11.4 or 12 (rounding up or giving a decimal answer).

Do not award an unevaluated calculation such as 80 ÷ 7 with no answer.

Example answers and how they were marked
Full marks1/1

11

Why: Correct: 80 ÷ 7 = 11 remainder 3, so 11 whole characters can be stored (1).

No marks0/1

11.4

Why: Near miss: gives a decimal result rather than the number of whole characters; the mark scheme does not accept 11.4.

No marks0/1

12

Why: Rounds up to 12, but 12 characters would need 84 bits, more than the 80 available; the mark scheme does not accept 12.

Questions by topic

More ways to revise Edexcel GCSE Computer Science (1CP2)

Teachers: build a paper, print it with the mark schemes at the end, and sign your school up to set your classes activities on every topic.

Fetch Decode Explain isn't affiliated with or endorsed by Pearson Edexcel.